text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Tags: brute force, constructive algorithms, greedy, implementation, strings
Correct Solution:
```
n,m = map(int,input().split())
dic = {}
dic2 = {}
for i in range(n):
s = input()
temp = list(s)
temp.reverse()
rev = "".join(temp)
if s == rev:
if s not in dic2:
dic2[s] = 0
dic2[s] += 1
elif s < rev:
if s not in dic:
dic[s] = [0,0]
dic[s][0] += 1
else:
if rev not in dic:
dic[rev] = [0,0]
dic[rev][1] += 1
lis = []
bis = []
for s in dic:
temp = list(s)
temp.reverse()
rev = "".join(temp)
for i in range(min(dic[s])):
lis.append(s)
bis.append(rev)
sub = ""
for s in dic2:
if dic2[s] % 2 == 1:
sub = s
for i in range(dic2[s] // 2):
lis.append(s)
bis.append(s)
bis.reverse()
ans = "".join(lis) +sub+ "".join(bis)
print (len(ans))
print (ans)
```
| 10,900 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
def ispal(s):
if (s==s[::-1]):
return 1
return 0
n,m=map(int,input().split())
A=[]
vis=[0]*n
for i in range(n):
A.append(list(input()))
if ispal(A[i]):
vis[i]=1
B=[]
C=[]
ans=0
for i in range(n):
if (vis[i]!=-1):
for j in range(i+1,n):
if (vis[j]!=-1 and A[i]==A[j][::-1]):
vis[i]=-1
vis[j]=-1
B.extend(A[i])
C.append(A[j])
ans=ans+2
break
for i in range(n):
if (vis[i]==1):
ans=ans+1
B.extend(A[i])
break
C.reverse()
for j in C:
B.extend(j)
print(len(B))
print(*B,sep="")
```
Yes
| 10,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
from collections import defaultdict,Counter
n,m = map(int,input().strip().split())
ls = []
for _ in range(n):
st = input()
ls.append(st)
start = ""
end = ""
temp = ""
ans = 0
dic = defaultdict(int)
f = 1
for i,val in enumerate(ls):
if i not in dic:
for j,value in enumerate(ls):
if j not in dic:
if i!=j and val==value[::-1]:
dic[i] = 1
dic[j] = 1
ans += 2*m
start += val
end = value+end
elif val==value[::-1] and f==1:
ans += m
temp = val
f = 0
print(ans)
if ans>0:
print(start+temp+end)
```
Yes
| 10,902 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
import math
t=1
while(t):
x,y=map(int,input().split())
l=[]
j=[]
k=[]
m=[]
t1=[]
t2=[]
for i in range(x):
l1=input()
l.append(l1)
for i in range(x):
r=l[:i]+l[i+1:]
if l[i][::-1] in r:
j.append(l[i])
k.append(l[i][::-1])
if(l[i]==l[i][::-1]):
m.append(l[i])
for i in range(len(j)):
if j[i] not in t1 and j[i][::-1] not in t1:
t1.append(j[i])
t2.append(j[i][::-1])
str1=str()
str2=str()
for i in range(len(t1)):
str1=str1+t1[i]
for i in range(len(t2)-1,-1,-1):
str2=str2+t2[i]
if(len(m)>0):
print(len(str1+m[0]+str2))
print(str1+m[0]+str2)
elif(len(str1+str2)>0):
print(len(str1+str2))
print(str1+str2)
else:
print(0)
t=t-1
```
Yes
| 10,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
n,m=map(int,input().split())
dp=[0]*n
for i in range(n):
s=input()
dp[i]=s
dpp=[]
dpop=[]
for i in range(n):
if dp[i]==dp[i][::-1]:
dpp.append(dp[i])
elif dp[i][::-1] in dp:
if dp[i][::-1] not in dpop:
dpop.append(dp[i])
dpop.append(dp[i][::-1])
s=""
if dpp:
s=dpp[0]
while dpop:
a=dpop.pop(0)
b=dpop.pop(0)
s=a+s+b
print(len(s))
print(s)
```
Yes
| 10,904 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
q, z = [int(x) for x in input().split()]
str1 = []
for x in range(q):
str1.append(input())
str2 = []
for each in str1:
str2.append(each[::-1])
str3 = set(str1) & set(str2)
'''
temp = set(str1) & set(str2)
temp1 = []
for each in str3:
temp.remove(each)
if each[::-1] not in temp:
temp1.append(each)
else:
temp.remove
for each in temp1:
str3.remove(each)
str4 = []
c = 0
for each in str3:
str4.insert(c, each)
str4.insert(-1*c-1, each[::-1])
str3.remove(each)
str3.remove(each[::-1])
num1 = len(str4) // 2
for each in temp1:
str4.insert(num1, each)
'''
str3 = list(str3)
#print(str3)
if len(str3) % 2 == 0:
for x in range(len(str3)):
if x == str3.index(str3[x][::-1]):
str3.pop(x)
break
for x in range(len(str3)):
temp = str3[x]
num1 = str3.index(temp[::-1])
if x != num1:
str3[num1], str3[-1*x-1] = str3[-1*x-1], str3[num1]
print(z*len(str3))
print("".join(str3))
```
No
| 10,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
n,m=map(int,input().split())
a=[]
ans=[]
res=''
for i in range(n):
a.append(input())
b=a.copy()
for i in a:
#print(a)
x=i
y=i[::-1]
a.remove(i)
#print(x,y)
if y in a:
ans.append(x)
a.remove(y)
else:
if x==y:
res=x
for i in a:
if i==i[::-1]:
res=i
break
s=''
if len(ans)==0:
if res!='':
print(len(res))
print(res)
else:
print(0)
print()
else:
for i in ans:
s+=i
if res!='':
s+=res+s[::-1]
else:
s+=s[::-1]
print(len(s))
print(s)
"""4 2
oo
ox
xo
xx4
"""
```
No
| 10,906 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
n, m = map(int, input().split())
s = []
c = ""
for i in range(n):
s.append(input())
while s:
try:
ind = s.index(a[::-1])
c = a + c + s[ind]
s.pop(ind)
except:
pass
s.pop(0)
print(len(c))
print(c)
```
No
| 10,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings "pop", "noon", "x", and "kkkkkk" are palindromes, while strings "moon", "tv", and "abab" are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don't print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
tabbat
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
baba
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
ababwxyzijjizyxwbaba
Note
In the first example, "battab" is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Submitted Solution:
```
n, m = map(int, input().split())
arr = []
pal = ''
rev = ''
for i in range(n):
temp = input()
arr.append(temp)
if temp == temp[::-1]:
rev = temp
for j in range(i):
if arr[i] == arr[j][::-1]:
pal = arr[i] + pal + arr[j]
if pal != '':
l = int((len(pal) + 1) / 2)
pal = pal[0:l] + rev + pal[l:]
if pal == '':
print(0)
else:
print(len(pal))
print(pal)
```
No
| 10,908 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
#!/usr/bin/env python
# coding: utf-8
# In[ ]:
def ter(n,x):
s=str(x)
a="1"
b="1"
maxi=False
for i in range(1,n):
if s[i]=="2":
if maxi:
a+="0"
b+="2"
else:
a+="1"
b+="1"
elif s[i]=="0":
a+="0"
b+="0"
else:
if maxi:
a+="0"
b+="1"
else:
maxi=True
a+="1"
b+="0"
print(int(a))
print(int(b))
t=int(input())
for i in range(t):
n=int(input())
x=int(input())
ter(n,x)
# In[ ]:
```
| 10,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
x = input()
a = x.find("1")
b = x.find("0")
d=""; e=""
if(a==-1):
for i in range(n):
if x[i]=="2":
d+="1"; e+="1"
else:
d+="0"; e+="0"
else:
for i in range(a):
if x[i]=="2":
d+="1"; e+="1"
else:
d+="0"; e+="0"
d+="1";e+="0"
d+="0"*(n-a-1)
e+=x[a+1:]
print(d+"\n"+e)
```
| 10,910 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for tt in range(t):
n = int(input())
x = input()
h1 = False
sol1 = ''
sol2 = ''
for xi in x:
if h1:
sol1 += '0'
sol2 += xi
else:
if xi == '0':
sol1 += '0'
sol2 += '0'
elif xi == '2':
sol1 += '1'
sol2 += '1'
else:
sol1 += '1'
sol2 += '0'
h1 = True
print(sol1)
print(sol2)
```
| 10,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
x = input()
a = []
b = []
c = 0
flag = False
for i in range(n):
if x[i]=='2':
a.append('1')
b.append('1')
elif x[i]=='0':
a.append('0')
b.append('0')
else:
a.append('1')
b.append('0')
c = i
flag = True
break
if flag==False or c==n-1:
a = ''.join(a)
b = ''.join(b)
else:
a = ''.join(a) + '0'*(n-c-1)
b = ''.join(b) + x[c+1:n]
print(a)
print(b)
```
| 10,912 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for k in range(t):
n = int(input())
x = input()
a, b = [], []
was_zero_one = False
for i, d in enumerate(x):
if d == '0':
a.append('0')
b.append('0')
elif d == '1':
if not was_zero_one:
a.append('1')
b.append('0')
was_zero_one = True
else:
a.append('0')
b.append('1')
elif d == '2':
if not was_zero_one:
a.append('1')
b.append('1')
else:
a.append('0')
b.append('2')
print("".join(a))
print("".join(b))
```
| 10,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
"""
Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
"""
def ternary_xor(s, n):
a, b = '', ''
flag = False
for i in range(n):
if s[i] == '2':
if flag:
a += '2'
b += '0'
else:
a += '1'
b += '1'
elif s[i] == '1':
a += '1'
b += '0'
if not flag:
flag = True
a, b = b, a
else:
a += '0'
b += '0'
if b > a:
a, b = b, a
print(a)
print(b)
if __name__== "__main__":
t = int(input())
for _ in range(t):
n = int(input())
s = input()
ternary_xor(s, n)
```
| 10,914 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
for h in range(int(input())):
n = int(input())
x = [int(i) for i in input()]
a = [0] * n
a[0] = 1
b = [0] * n
b[0] = 1
fl = 0
for i in range(1, n):
if fl:
b[i] = x[i]
elif x[i] == 1:
a[i] = 1
fl = 1
elif x[i] == 2:
a[i] = 1
b[i] = 1
[print(i, end="") for i in a]
print()
[print(i, end="") for i in b]
print()
```
| 10,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
a, b = '', ''
is_one_found = False
for x in s:
if is_one_found:
a += '0'
b += x
else:
num = int(x)
if num == 1:
is_one_found = True
a += x
b += '0'
else:
a += str(num // 2)
b += str(num // 2)
print(a, b, sep='\n')
```
| 10,916 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
lst = list(input())
a, b = ['0']*n, ['0']*n
for i in range(n):
if lst[i] == '1':
a[i] = '1'
b[i] = '0'
for j in range(i+1, n):
b[j] = lst[j]
break
elif lst[i] == '0':
a[i] = b[i] = '0'
elif lst[i] == '2':
a[i] = b[i] = '1'
print(''.join(a))
print(''.join(b))
```
Yes
| 10,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
lis = input()
s1 = []
s2 = []
diff = 0
for i in lis:
if diff == 0:
if i == '2':
s1.append(1)
s2.append(1)
elif i == '0':
s1.append(0)
s2.append(0)
else:
s1.append(1)
s2.append(0)
diff = 1
else:
s1.append(0)
s2.append(i)
for i in range(len(s1)):
print(s1[i],end='')
print()
for i in range(len(s1)):
print(s2[i],end='')
print()
```
Yes
| 10,918 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
from collections import Counter
from collections import defaultdict
import math
t=int(input())
for _ in range(0,t):
lis,mis=list(),list()
n =input()
n=int(n)
x =input()
x=list(x)
x=list(map(int,x))
s1=x[0]
if(s1 > 1):
mis.append(1)
lis.append(1)
g=1
i = g
while(i < n):
if(x[i] == 0):
lis.append(0)
mis.append(0)
elif(x[i] == 2):
lis.append(1)
mis.append(1)
elif(x[i] == 1):
lis.append(1)
mis.append(0)
i += 1
break
i=i+1
while(i<n):
k=0
lis.append(k)
mis.append(x[i])
i=i+ 1
lis=map(str,lis)
mis=map(str,mis)
s1="".join(list(lis))
s2="".join(list(mis))
print(s1)
print(s2)
```
Yes
| 10,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
T = int(input())
while T > 0 :
n = int(input())
val = input()
a = "1"
b = "1"
count = 1
for x in range(1, n) :
if val[x] == "2" :
a = a + "1"
b = b + "1"
elif val[x] == "0" :
a = a + "0"
b = b + "0"
else :
break
count += 1
if count < n :
if val[count] == "1" :
a = a + "1"
b = b + "0"
count = count + 1
for x in range(count, n) :
if val[x] == "2" :
a = a + "0"
b = b + "2"
elif val[x] == "1" :
a = a + "0"
b = b + "1"
else :
a = a + "0"
b = b + "0"
print(a)
print(b)
T -= 1
```
Yes
| 10,920 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
try:
t=int(input())
for _ in range(t):
a=''
b=''
n=int(input())
k=input()
for i in range(n):
if(k[i]=='1'):
a+='1'
b+='0'
for j in range(i+1,n):
a+='0'
b+=k[i]
break
else:
if(k[i]=='2'):
a+='1'
b+='1'
else:
a+='0'
b+='0'
print(a)
print(b)
except Exception:
pass
```
No
| 10,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
q = int(input())
for _ in range(q):
input()
x = input()
ans1 = ['1']
ans2 = ['1']
is_one = False
for i in x[1:]:
if i == '0':
ans1.append('0')
ans2.append('0')
elif i == '1':
if is_one:
ans1.append('0')
ans2.append('1')
else:
ans1.append('1')
ans2.append('1')
is_one = True
else:
if is_one:
ans1.append('0')
ans2.append('2')
else:
ans1.append('1')
ans2.append('1')
for i in ans1:
print(i, end='')
print()
for i in ans2:
print(i, end='')
print()
```
No
| 10,922 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
cases = int(input())
def ternaryXOR( value):
value = [x for x in value]
a = []
b = []
for x in value:
if x == "2":
a.append("1")
b.append("1")
elif x == "1":
if int("".join(a)) < int("".join(b)):
a.append("0")
b.append("1")
else:
a.append("1")
b.append("0")
else:
a.append("0")
b.append("0")
return "".join(a), "".join(b)
for _ in range(cases):
input()
value = input()
a, b = ternaryXOR(value)
print(a)
print(b)
```
No
| 10,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.
You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.
Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.
Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1 ≤ n ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).
Output
For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.
Example
Input
4
5
22222
5
21211
1
2
9
220222021
Output
11111
11111
11000
10211
1
1
110111011
110111010
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
x = list(map(int, list(input())))
a = [1]
b = [1]
for i in range(1, n):
if x[i] == 2:
a.append(1)
b.append(1)
elif x[i] == 1:
a.append(1)
b.append(0)
else:
a.append(0)
b.append(0)
print("".join(map(str, a)))
print("".join(map(str, b)))
```
No
| 10,924 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
import math
cases = int(input())
for t in range(cases):
n,k = list(map(int,input().split()))
s = ''.join(sorted(input()))
if k==1:
print(''.join(s))
elif s[0]!=s[k-1]:
print(s[k-1])
else:
if s[0]==s[-1]:
print(s[0]*(math.ceil(n/k)))
elif s[k]==s[-1]:
print(s[0]+s[k]*math.ceil((n-k)/k))
else:
print(s[0]+s[k:])
```
| 10,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
given = sorted(input())
if len(set(given[:k]))!=1:
ans = given[k-1]
else:
ans = ''
if len(set(given[k:]))==1:
for i in range(0,n,k):
ans+=given[i]
else:
ans = ''.join(given[k-1:])
print(ans)
```
| 10,926 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
# cook your dish here
t=int(input())
for _ in range(t):
n,k=map(int,input().split())
s=list(input())
s.sort()
c=s[0]
if s[k-1]!=c:
print(s[k-1])
else:
x=set(s[k:])
if len(x)==0 or len(x)==1:
ans=['']*k
j=0
for i in range(n):
ans[j]=ans[j]+s[i]
j=(j+1)%k
ans.sort()
m=ans[-1]
else:
m=''
for i in range(k-1,n):
m=m+s[i]
print(m)
```
| 10,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
for lo in range(int(input())):
#n = int(input())
n,k = map(int,input().split())
st = input()
ls = [0 for i in range(26)]
c = 0
mn = 30
for i in st:
x = ord(i)-97
if ls[x]==0:
c+=1
ls[x]+=1
mn = min(x,mn)
if c==1:
z = 0
if ls[mn]%k!=0:
z+=1
z+=(ls[mn]//k)
ans = ""
for i in range(z):
ans+=chr(97+mn)
print(ans)
continue
if ls[mn]<k:
z = 0
for i in range(26):
z+=ls[i]
if mn!=i and z>=k:
ans = chr(97+i)
break
print(ans)
continue
if ls[mn]==k and c==2:
x = 0
mn2 = 0
for i in range(26):
if mn!=i and ls[i]>0:
mn2 = i
x = ls[i]
break
z = 0
if ls[mn2]%k!=0:
z+=1
z+=(ls[mn2]//k)
ans = chr(97+mn)
for i in range(z):
ans+=chr(97+mn2)
print(ans)
continue
z = 0
ans = ""
for i in range(26):
if mn==i:
for j in range(ls[mn]-k+1):
ans+=chr(mn+97)
else:
for j in range(ls[i]):
ans+=chr(i+97)
print(ans)
```
| 10,928 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int, input().split())
s = ''.join(sorted(input()))
if(s[0] != s[k-1] or k == n):
print(s[k-1])
continue
if(s[k] != s[n-1]):
print(s[0]+s[k:])
else:
print(s[0]+s[n-1]*((n-1)//k))
```
| 10,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
import math
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
s = input()
s1 = [i for i in s]
s1.sort()
s = ''.join(s1)
#print(s)
if m==n:
print(s[n-1])
continue
if m==1:
print(s)
continue
lis = [[s[i]] for i in range(m)]
pos = -1
for i in range(m-1):
if lis[i]==lis[m-1]:
pos = i
break
if pos==-1:
print(*lis[m-1])
continue
pp = pow(2,10)
for i in range(20):
pp+=i
if pos==0:
if s[m]==s[n-1]:
print(s[0]+s[m]*(math.ceil((n-m)/m)))
else:
print(s[0]+s[m:])
else:
print(*lis[m-1])
```
| 10,930 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
I=input
exec(int(I())*"n,k=map(int,I().split());s=''.join(sorted(I()));c=s[k-1];print(c+(c==s[0])*s[k::k**(s[k%n]==s[-1])]);")
```
| 10,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
#ライブラリインポート
from collections import defaultdict
con = 10 ** 9 + 7
#入力受け取り
def getlist():
return list(map(int, input().split()))
#処理内容
def main():
T = int(input())
for i in range(T):
N, K = getlist()
s = sorted(list(input()))
#面倒
if N == K:
print(s[-1])
elif s[0] != s[K - 1]:
print(s[K - 1])
elif s[0] == s[-1]:
var = 0
if N % K == 0:
var = int(N // K)
else:
var = int(N // K) + 1
ans = s[0] * var
print(ans)
elif s[K] == s[-1]:
ans = s[0]
var = 0
if (N - K) % K == 0:
var = int(N // K) - 1
else:
var = int(N // K)
ans += var * s[-1]
print(ans)
else:
ans = s[0]
for j in range(K, N):
ans += s[j]
print(ans)
if __name__ == '__main__':
main()
```
| 10,932 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
import math
t=int(input())
def als(A):
if len(list(set(list(A))))==1:
return True
return False
for _ in range(t):
n,k=list(map(int,input().split()))
s=input()
ss=s
ss=list(ss)
ss.sort()
ss=''.join(ss)
bb=list(set(list(ss)))
bb=''.join(bb)
a=[0]*26
for x in s:
a[ord(x)-97]+=1
tt=[]
idx=[]
for i in range(26):
if a[i]!=0:
tt.append(a[i])
idx.append(i)
if tt[0]<k:
print(ss[k-1])
elif als(ss[k:]):
res=ss[0]
val=math.ceil(len(ss[k:])/k)*ss[k]
print(res+val)
else:
print(ss[0]+ss[k:])
```
Yes
| 10,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
for u in range(int(input())):
n,k=map(int,input().split())
s=''.join(sorted(input()))
if(k==n or s[0]!=s[k-1]):
print(s[k-1])
elif(s[k]==s[-1]):
print(s[0]+s[k]*((n-1)//k))
else:
print(s[k-1:])
```
Yes
| 10,934 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
import sys
#from math import *
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output2.txt','w')
t=int(input())
while t>0:
t-=1
n,k=map(int,input().split())
s=list(input().rstrip())
s.sort()
d={}
a=["" for i in range(k)]
for i in range(k):
a[i%k]+=s[i]
j=0
i=k
if len(set(s[k:]))==1:
for i in range(k,len(s)):
a[j%k]+=s[i]
j+=1
if j==k or a[j][0]>a[0][0]:
j=0
else:
for i in range(k,len(s)):
a[j%k]+=s[i]
#print(a)
print(max(a))
```
Yes
| 10,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
s = sorted([c for c in input()])
if s[k-1] != s[0]:
print(s[k-1])
else:
fi = s[0]
se = None
three = False
for i in range(n):
if s[i] != fi:
if se == None:
se = s[i]
elif s[i] != se:
three = True
break
if three:
print("".join(s[k-1:]))
else:
i = 0
while i < n and s[0] == s[i]:
i += 1
if i == n:
string = [s[0]] * (n//k)
if n % k != 0:
string += [s[0]]
print("".join(string))
elif i == k:
if (n - i) % k == 0:
print(s[0] + s[i]*((n-i)//k))
else:
print(s[0] + s[i]*((n-i)//k + 1))
else:
print("".join(s[k-1:]))
```
Yes
| 10,936 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return stdin.read().split()
range = xrange # not for python 3.0+
inp=inp()
pos=1
for t in range(int(inp[0])):
n,k=int(inp[pos]),int(inp[pos+1])
pos+=2
s=list(inp[pos])
pos+=1
s.sort()
ans=[s[i] for i in range(k)]
if n==k or ans[0]!=ans[-1]:
pr(ans[-1]+'\n')
continue
if s[k]==s[-1]:
ln=n-k
pr(ans[0]+''.join(s[k]*((ln/k)+int(ln%k!=0)))+'\n')
else:
pr(ans[0]+''.join(s[k:])+'\n')
```
Yes
| 10,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
import sys,os.path
import math
if __name__ == '__main__':
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(int(input())):
n,k = map(int,input().split())
s = input()
l = [0 for i in range(26)]
flag = True
for i in range(n):
l[ord(s[i])-ord('a')]+=1
for i in range(n-1):
if s[i]!=s[i+1]:
flag = False
break
if flag:
ans = s[0]*(math.ceil(n/k))
print(ans)
else:
f = False
first = 26
for i in range(26):
if l[i]!=0:
first = min(first,i)
if l[i]!=0 and l[i]!=k:
f = True
break
if l[first]<k:
su = 0
for i in range(26):
su+=l[i]
if su>=k:
print(chr(97+i))
break
else:
if not f:
ans = ""
val = math.ceil(l[first]/k)
a = chr(97+first)*val
ans+=a
for i in range(first+1,26):
if l[i]!=0:
val = math.ceil(l[i]/k)
ans+=chr(97+i)*val
else:
ans = ""
ans += chr(97+first)*(l[first]-k+1)
for i in range(first+1,26):
if l[i]!=0:
a = chr(97+i)*l[i]
ans+=a
print(ans)
```
No
| 10,938 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
a = sorted(input())
ans = a[k-1]
if(len(set(a[0:k]))>1):
print(ans)
else:
for i in range(k, n):
if(len(set(a[i:]))>1):
ans = ans + a[i]
else :
ans = ans + a[i]
break
print(ans)
```
No
| 10,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
import sys
try:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
for _ in range(int(input())):
n,k = [int(x) for x in input().split()]
s = str(input())
s = "".join(sorted(s))
if k==n: print(s[n-1]); continue
if s[0]!=s[k-1]: print(s[k-1]); continue
print(s[0],end="")
if s[k]==s[n-1]:
print(k,n-1)
for _ in range(int((n-k)/k)): print(s[k],end="")
if (n-k)%k: print(s[k],end="")
print()
else:
print(s[k:])
```
No
| 10,940 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.
For example, if s = baba and k=2, Phoenix may distribute the letters of his string in many ways, such as:
* ba and ba
* a and abb
* ab and ab
* aa and bb
But these ways are invalid:
* baa and ba
* b and ba
* baba and empty string (a_i should be non-empty)
Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).
String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.
The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.
The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.
It is guaranteed that the sum of n over all test cases is ≤ 10^5.
Output
Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.
Example
Input
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
Output
ab
abbc
b
aa
x
ehinopx
Note
In the first test case, one optimal solution is to distribute baba into ab and ab.
In the second test case, one optimal solution is to distribute baacb into abbc and a.
In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.
In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.
In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.
In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
Submitted Solution:
```
def solve():
N, K = [int(el) for el in input().split()]
s = input()
arr = ['' for _ in range(K)]
ind = -1
for ch in sorted(s):
ind = (ind + 1) % K
arr[ind] += ch
arr.sort()
return arr[-1]
for t in range(1, int(input()) + 1):
print(solve())
```
No
| 10,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
text = int(input())
for i in range(text):
a = int(input())
if a % 4==0:
print('YES')
else:
print('NO')
```
| 10,942 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
:license: GPLv3 --- Copyright (C) 2020 Olivier Pirson
:author: Olivier Pirson --- http://www.opimedia.be/
"""
import sys
#
# Main
######
def main():
"""Run main work."""
nb = int(sys.stdin.readline())
for _ in range(nb):
n = int(sys.stdin.readline())
print('YES' if n % 4 == 0
else 'NO')
if __name__ == '__main__':
main()
```
| 10,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
from collections import defaultdict as dd
import math
import sys
import string
input=sys.stdin.readline
def nn():
return int(input())
def li():
return list(input())
def mi():
return map(int, input().split())
def lm():
return list(map(int, input().split()))
q=nn()
for _ in range(q):
n = nn()
if n%4==0:
print("YES")
else:
print("NO")
```
| 10,944 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
if(not n%4):print("Yes")
else:print("No")
```
| 10,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
T=int(input())
for _ in range(T):
N=int(input())
if N%4==0:
print("YES")
else:
print("NO")
```
| 10,946 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from math import floor
from bisect import bisect_right
from collections import Counter
from math import gcd
mod=998244353
def main():
for _ in range(int(input())):
n=int(input())
# k=180*(n-2)//n
if n%4==0:
print('YES')
else:
print('NO')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
| 10,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
if n % 4 == 0:
print("YES")
else:
print("NO")
```
| 10,948 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Tags: geometry, math
Correct Solution:
```
tc = int(input())
for i in range(tc):
N = int(input())
if N%4 == 0:
print("yEs")
else:
print("nO")
if N == 69420:
print("bing bong")
```
| 10,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
for _ in range(int(input())):
p = int(input())
if p %4 != 0 :
print('NO')
else:
print('YES')
```
Yes
| 10,950 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
t = int(input())
# print('t', t)
for _ in range(t):
n = int(input())
# print('n', n)
if n%4==0:
print('YES')
else:
print('NO')
```
Yes
| 10,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
import sys, math
if __name__ == "__main__":
t = int(input())
for i in range(t):
ni = int(input())
if ni % 4 == 0:
print("YES")
else:
print("NO")
```
Yes
| 10,952 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
for j in range(int(input())):
n=int(input())
if n%4!=0:
print("NO")
else:
print("YES")
```
Yes
| 10,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
import math
import datetime
import collections
import statistics
import itertools
def is_prime(num):
for i in range(2, int(math.sqrt(num)) + 1):
if num % i == 0:
return False
return True
def input_list():
ll = list(map(int, input().split(" ")))
return ll
tc = int(input())
for _ in range(tc):
n = int(input())
if n % 2 == 0:
print("YES")
else:
print("NO")
```
No
| 10,954 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
if(n==3):
print("NO")
else:
print("YES")
```
No
| 10,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
def f():
n=int(input(""))
if n>3:
print("YES")
else:
print("NO")
n=int(input(""))
for i in range(n):
f()
```
No
| 10,956 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lee is going to fashionably decorate his house for a party, using some regular convex polygons...
Lee thinks a regular n-sided (convex) polygon is beautiful if and only if he can rotate it in such a way that at least one of its edges is parallel to the OX-axis and at least one of its edges is parallel to the OY-axis at the same time.
Recall that a regular n-sided polygon is a convex polygon with n vertices such that all the edges and angles are equal.
Now he is shopping: the market has t regular polygons. For each of them print YES if it is beautiful and NO otherwise.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of polygons in the market.
Each of the next t lines contains a single integer n_i (3 ≤ n_i ≤ 10^9): it means that the i-th polygon is a regular n_i-sided polygon.
Output
For each polygon, print YES if it's beautiful or NO otherwise (case insensitive).
Example
Input
4
3
4
12
1000000000
Output
NO
YES
YES
YES
Note
In the example, there are 4 polygons in the market. It's easy to see that an equilateral triangle (a regular 3-sided polygon) is not beautiful, a square (a regular 4-sided polygon) is beautiful and a regular 12-sided polygon (is shown below) is beautiful as well.
<image>
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
def solve(N, ):
if N % 2 == 0:
return "YES"
return "NO"
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
T = int(input())
for t in range(T):
N, = [int(x) for x in input().split()]
ans = solve(N, )
print(ans)
```
No
| 10,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
import sys
import operator
import array
#-----------
def solve():
a = [int(x) for x in input().split()]
n = int(input())
b = [int(x) for x in input().split()]
have = [ [] ] * (n*6)
arr_append = have.append
for i in range(0, n):
for j in range(0, 6):
#arr_append([ b[i] - a[j], i ])
have[i*6 + j] = [ b[i] - a[j], i ]
cnt = array.array('L', [0])*n
z = n
sz = len(have)
ans = 999999999999
r = 0
have.sort(key=operator.itemgetter(0))
for i in range(0, sz):
while (r < sz) and (z > 0):
cnt[have[r][1]] += 1
if (cnt[have[r][1]] == 1):
z-=1
r+=1
if (z > 0):
break
ans = min(ans, have[r - 1][0] - have[i][0]);
cnt[have[i][1]] -= 1
if (cnt[have[i][1]] == 0):
z+=1
print(ans)
#-----------
def main(argv):
solve()
if __name__ == "__main__":
main(sys.argv)
```
| 10,958 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# ------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# sys.setrecursionlimit(300000)
# from heapq import *
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow
import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
# from functools import lru_cache
a = RLL()
a.sort()
n = N()
b = RLL()
data = [(b[i]-a[j],i) for i in range(n) for j in range(6)]
# print(data)
data.sort()
res = float('inf')
now = [0]*n
count = 0
l,r = 0,0
mn = 6*n
while 1:
while count<n and r<mn:
k = data[r][1]
now[k]+=1
if now[k]==1:
count+=1
r+=1
if count<n:
break
while count==n:
k = data[l][1]
now[k]-=1
if now[k]==0:
count-=1
l+=1
res = min(data[r-1][0]-data[l-1][0],res)
print(res)
```
| 10,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin
input = stdin.readline
a = sorted([int(i) for i in input().split()])
n = int(input())
b = sorted([int(i) for i in input().split()])
c = []
for i in range(n):
c += [[b[i] - a[j], i] for j in range(6)]
c.sort()
d = [0] * n
e = 0
ans = 10 ** 10
u = 0
for i in range(len(c)):
while u < len(c) and e < n:
x = c[u][1]
if d[x] == 0:
e += 1
d[x] += 1
u += 1
if e == n:
ans = min(ans, c[u - 1][0] - c[i][0])
x = c[i][1]
d[x] -= 1
if d[x] == 0:
e -= 1
print(ans)
```
| 10,960 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
a=list(map(int,input().split()));n=int(input());s=list(map(int,input().split()));b=[];i=j=0;ans=10**18;cs=[0]*n;nz=1;z=n*6
for y in range(n):
for x in a:b.append((s[y]-x)*n+y)
b.sort();cs[b[0]%n]+=1
while j+1<z:
while j+1<z and nz<n:j+=1;nz+=cs[b[j]%n]<1;cs[b[j]%n]+=1
while nz==n:ans=min(ans,b[j]//n-b[i]//n);cs[b[i]%n]-=1;nz-=cs[b[i]%n]==0;i+=1
print(ans)
```
| 10,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
#If FastIO not needed, used this and don't forget to strip
import sys
input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
"""
#from collections import Counter as cc
#import math, string
def getInts():
return [int(s) for s in input().strip().split()]
def getInt():
return int(input().strip())
"""
Fret = B[j] minus A[i]
There are 6 possible values for each note
1 4 8 13 20 25
1 4 9 13 20 25
0 0 1 0 0 0
Order the notes
1,2,3,4,5,6
(0,-1,-1,-1,-1,-1)
(1,0,-1,-1,-1,-1)
(1,2,3,4,5,6)
Binary search? Is it possible to attain a minimum difference of <= D?
We need to find a range [L,R] within which every note can be played, such that R-L is minimal
"""
def solve():
A = getInts()
M = getInt()
B = getInts()
X = []
P = []
for i, b in enumerate(B):
for a in A:
P.append((b-a,i))
P.sort()
i = 0
j = -1
counts = [0]*M
sset = set()
ans = 2*10**9
MAX = M*6
set_len = 0
while i < MAX:
while set_len < M and j < MAX:
j += 1
try:
if not counts[P[j][1]]: set_len += 1
counts[P[j][1]] += 1
except:
break
if set_len < M:
break
z = P[i][1]
ans = min(ans,P[j][0]-P[i][0])
counts[z] -= 1
if not counts[z]:
set_len -= 1
i += 1
return ans
#for _ in range(getInt()):
print(solve())
```
| 10,962 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**32, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:max(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
a=list(map(int,input().split()))
n=int(input())
l=list(map(int,input().split()))
a.sort()
l.sort()
d=[]
for i in range(n):
for j in range(6):
d.append((l[i]-a[j],i))
t=[0]*n
d.sort()
st=0
end=0
cur=[0]*n
s=SegmentTree1(cur)
ans=-1
while(end<len(d)):
cur[d[end][1]]+=1
s.__setitem__(d[end][1],cur[d[end][1]])
if s.query(0,n-1)==0:
end+=1
else:
if ans==-1:
ans=d[end][0]-d[st][0]
ans=min(ans,d[end][0]-d[st][0])
cur[d[st][1]]-=1
s.__setitem__(d[st][1], cur[d[st][1]])
cur[d[end][1]]-=1
st+=1
if st>end:
end=st
print(ans)
```
| 10,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
def main():
a=[int(x) for x in input().split()]
n=int(input())
b=[int(x) for x in input().split()]
fretsAndNoteNo=[] #(fret number, note number)
for base in a:
for noteNo,note in enumerate(b):
fret=note-base
fretsAndNoteNo.append((fret,noteNo))
fretsAndNoteNo.sort(key=lambda x:x[0]) #sort by fret number ascending
ans=float('inf')
noteNoCnts=[0 for _ in range(n)]
totalUniqueNotesInSegment=0
left=0
for fretNo,noteNo in fretsAndNoteNo:
noteNoCnts[noteNo]+=1
if noteNoCnts[noteNo]==1:
totalUniqueNotesInSegment+=1
while totalUniqueNotesInSegment==n: #keep moving the left pointer
diff=fretNo-fretsAndNoteNo[left][0]
ans=min(ans,diff)
noteNoCnts[fretsAndNoteNo[left][1]]-=1
if noteNoCnts[fretsAndNoteNo[left][1]]==0:
totalUniqueNotesInSegment-=1
left+=1
print(ans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
main()
```
| 10,964 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
import sys
import operator
import array
#-----------
def solve():
a = (list(map(int, input().split())))
n = int(input())
b = (list(map(int, input().split())))
have = [ [] ] * (n*6)
#have = []
arr_append = have.append
c=0
for i in range(0, n):
for j in range(0, 6):
#arr_append(( b[i] - a[j], i ))
have[c] = ( b[i] - a[j], i )
c+=1
have.sort(key=operator.itemgetter(0))
cnt = array.array('L', [0])*n
z = n
sz = len(have)
ans = 999999999999
r = 0
for i in range(0, sz):
while (r < sz) and (z > 0):
cnt[have[r][1]] += 1
if (cnt[have[r][1]] == 1):
z-=1
r+=1
if (z > 0):
break
ans = min(ans, have[r - 1][0] - have[i][0]);
cnt[have[i][1]] -= 1
if (cnt[have[i][1]] == 0):
z+=1
print(ans)
#-----------
def main(argv):
solve()
if __name__ == "__main__":
main(sys.argv)
```
| 10,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
import sys
import heapq
input = sys.stdin.readline
a = list(set(map(int, input().split())))
k = len(a)
n = int(input())
b = list(set(map(int, input().split())))
n = len(b)
a.sort()
b.sort()
c = [0 for i in range(n)]
mn = float("inf")
pq = []
for i in range(n):
heapq.heappush(pq, (- b[i] + a[c[i]], i))
mn = b[0] - a[0]
x = float("inf")
while True:
td = heapq.heappop(pq)
i = td[1]
x = min(x, -td[0] - mn)
c[i] += 1
if c[-1] == k:
break
mn = min(b[i] - a[c[i]], mn)
heapq.heappush(pq, (- b[i] + a[c[i]], i))
print(x)
```
Yes
| 10,966 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
def main():
a=[int(x) for x in input().split()]
n=int(input())
b=[int(x) for x in input().split()]
fretsAndNoteNo=[] #(fret number, note number)
for base in a:
for noteNo,note in enumerate(b):
fret=note-base
fretsAndNoteNo.append((fret,noteNo))
fretsAndNoteNo.sort(key=lambda x:x[0]) #sort by fret number ascending
ans=float('inf')
noteNoCnts=[0 for _ in range(n)]
totalUniqueNotesInSegment=0
left=0
for right,(fretNo,noteNo) in enumerate(fretsAndNoteNo):
noteNoCnts[noteNo]+=1
if noteNoCnts[noteNo]==1:
totalUniqueNotesInSegment+=1
while totalUniqueNotesInSegment==n: #keep moving the left pointer
diff=fretNo-fretsAndNoteNo[left][0]
ans=min(ans,diff)
noteNoCnts[fretsAndNoteNo[left][1]]-=1
if noteNoCnts[fretsAndNoteNo[left][1]]==0:
totalUniqueNotesInSegment-=1
left+=1
print(ans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
main()
```
Yes
| 10,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
from itertools import chain as _chain
import sys as _sys
def main():
a_seq = tuple(_read_ints())
n, = _read_ints()
notes = tuple(_read_ints())
result = find_minimal_melody_difficulty(a_seq, notes)
print(result)
def find_minimal_melody_difficulty(strings_values, notes):
strings_values = set(strings_values)
strings_values = sorted(strings_values, reverse=True)
if len(strings_values) == 1:
return max(notes) - min(notes)
frets_by_notes_indices = [[note - x for x in strings_values] for note in notes]
# frets_by_notes_indices[i_note] is sorted for every correct i_note
del strings_values
del notes
max_initially_selected_fret = max(frets[0] for frets in frets_by_notes_indices)
for frets in frets_by_notes_indices:
i_first_interesting_fret = 0
while i_first_interesting_fret + 1 < len(frets) \
and frets[i_first_interesting_fret+1] <= max_initially_selected_fret:
i_first_interesting_fret += 1
frets[:] = frets[i_first_interesting_fret:]
sorted_frets = sorted(_chain.from_iterable(frets_by_notes_indices))
initially_selected_frets = [frets[0] for frets in frets_by_notes_indices]
min_initially_selected = min(initially_selected_frets)
i_first_selected = sorted_frets.index(min_initially_selected)
while i_first_selected+1 < len(sorted_frets) \
and sorted_frets[i_first_selected+1] == min_initially_selected:
i_first_selected += 1
i_first_selected -= initially_selected_frets.count(min_initially_selected) - 1
i_last_selected = i_first_selected + len(initially_selected_frets) - 1
available_selections = _chain.from_iterable(
zip(frets[1:], frets[0:])
for frets in frets_by_notes_indices
)
available_selections = sorted(available_selections)
# TODO: can replace it with indices_to_unselect
# defaultdict(int) is slower
frets_to_unselect = dict()
for frets in frets_by_notes_indices:
for fret in frets:
frets_to_unselect[fret] = 0
result = sorted_frets[i_last_selected] - sorted_frets[i_first_selected]
for new_fret, fret_to_unselect in available_selections:
frets_to_unselect[fret_to_unselect] += 1
i_first_selected = _shift_index_skipping(sorted_frets, i_first_selected, frets_to_unselect)
i_last_selected += 1
current_maxmin_difference = sorted_frets[i_last_selected] - sorted_frets[i_first_selected]
if current_maxmin_difference < result:
result = current_maxmin_difference
return result
def _shift_index_skipping(sorted_seq, index, deletions_ns):
while index < len(sorted_seq) and deletions_ns[sorted_seq[index]] > 0:
deleted_element = sorted_seq[index]
index += 1
deletions_ns[deleted_element] -= 1
return index
def _read_line():
result = _sys.stdin.readline()
assert result[-1] == "\n"
return result[:-1]
def _read_ints():
return map(int, _read_line().split())
if __name__ == '__main__':
main()
```
Yes
| 10,968 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
def solve(A, N, B):
costs = []
indices = []
for i, a in enumerate(A):
for j, b in enumerate(B):
costs.append(b - a)
indices.append(j)
order = list(range(len(costs)))
order.sort(key=lambda i: costs[i])
costs2 = []
indices2 = []
for i in order:
costs2.append(costs[i])
indices2.append(indices[i])
costs = costs2
indices = indices2
window = Counter() # indices covered by this window
windowC = deque()
windowI = deque()
tail = 0
best = float("inf")
for i in range(len(costs)):
b = costs[i]
j = indices[i]
while len(window) != N and tail < len(costs):
bb = costs[tail]
jj = indices[tail]
window[jj] += 1
windowC.append(bb)
windowI.append(jj)
tail += 1
if len(window) != N:
break
best = min(best, windowC[-1] - windowC[0])
bb = windowC.popleft()
jj = windowI.popleft()
window[jj] -= 1
if window[jj] == 0:
del window[jj]
return best
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
A = [int(x) for x in input().split()]
(N,) = [int(x) for x in input().split()]
B = [int(x) for x in input().split()]
ans = solve(A, N, B)
print(ans)
```
Yes
| 10,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
# region fastio # from https://codeforces.com/contest/1333/submission/75948789
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#endregion
from heapq import heappush, heappop
A = sorted(map(int, input().split()), reverse=True)
N = int(input())
q = []
B = sorted(map(int, input().split()))
ma = B[-1] - A[0]
for b in B:
ba = b-A[0] # fret
q.append(ba<<4 | 0)
ans = 1<<62
while True:
v = heappop(q)
ba = v >> 4
idx_A = v & 0b1111
mi = ba
an = ma - mi
if an < ans:
ans = an
if idx_A == 5:
break
ba += A[idx_A]
ba -= A[idx_A+1]
heappush(q, ba<<4|idx_A+1)
print(ans)
```
No
| 10,970 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import Counter as cc
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
Fret = B[j] minus A[i]
There are 6 possible values for each note
1 4 8 13 20 25
1 4 9 13 20 25
0 0 1 0 0 0
Order the notes
1,2,3,4,5,6
(0,-1,-1,-1,-1,-1)
(1,0,-1,-1,-1,-1)
(1,2,3,4,5,6)
Binary search? Is it possible to attain a minimum difference of <= D?
We need to find a range [L,R] within which every note can be played, such that R-L is minimal
"""
def solve():
N = 6
A = getInts()
M = getInt()
B = getInts()
X = []
for i in range(M):
tmp = []
for j in range(N):
tmp.append(B[i]-A[j])
tmp.sort()
X.append(tmp)
P = []
for i in range(M):
for j in range(N):
P.append((X[i][j],i))
P.sort()
i = 0
j = -1
counts = cc()
sset = set()
ans = 2*10**9
while i < M*N:
while len(sset) < M and j < M*N:
j += 1
try:
sset.add(P[j][1])
except:
break
if j == M*N:
break
z = P[i][1]
ans = min(ans,P[j][0]-P[i][0])
counts[z] -= 1
if not counts[z]:
sset.remove(z)
i += 1
return ans
#for _ in range(getInt()):
print(solve())
```
No
| 10,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
aa = sorted(set(map(int, input().split())))
n = int(input())
bb = sorted(map(int, input().split()))
inf = float('inf')
if n == 1:
print(0)
if len(aa) == 1:
print(bb[-1] - bb[0])
else:
ans = inf
for ca in aa:
rr = minr = -inf
ll = maxl = inf
c = bb[0] - ca
for b in bb[1:]:
mfc = bminr = inf
bmaxl = -inf
for a in aa:
f = b - a
if f == c:
bminr = bmaxl = brr = bll = c
fc = abs(f - c)
if fc < mfc:
mfc = fc
if f > c:
brr = f
bll = None
else:
brr = None
bll = f
if f > c:
bminr = min(bminr, f)
else:
bmaxl = max(bmaxl, f)
minr = max(minr, bminr)
maxl = min(maxl, bmaxl)
if bll is not None:
ll = min(ll, bll)
if brr is not None:
rr = max(rr, brr)
ans = min(ans, abs(minr - c), abs(c - maxl), abs(rr - ll))
print(ans)
```
No
| 10,972 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image>
Submitted Solution:
```
z=list(map(int,input().split()))
r=int(input())
sr=list(map(int,input().split()))
maxa=max(sr)-max(z)
ans=[]
import math
for i in range(len(sr)):
am=math.inf
for j in range(6):
if(sr[i]-z[j]-maxa>0):
continue;
am=min(am,abs((sr[i]-z[j])-maxa))
ans.append(am)
t1=max(ans)
mini=min(sr)-min(z)
ans=[]
for i in range(len(sr)):
am=math.inf
for j in range(6):
if(sr[i]-z[j]<mini):
continue;
else:
am=min(am,sr[i]-z[j]-mini)
ans.append(am)
t2=max(ans)
print(min(t1,t2))
```
No
| 10,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
for t in range(int(input())):
x,y=map(int, input().split())
print(x-1,y,end=' ')
```
| 10,974 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
from sys import stdin
input=stdin.readline
for _ in range(int(input())):
a,b=map(int,input().split())
print(a-1,b)
```
| 10,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
mod = 10**9 + 7
def solve():
x, y = map(int, input().split())
print(x - 1, y)
t = 1
t = int(input())
while t > 0:
solve()
t -= 1
```
| 10,976 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
for i in range(int(input())):
x,y=map(int,input().split())
print(x-1,y,end=" ")
print()
```
| 10,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
t = int(input())
for i in range(t):
n, m = map(int, input().split())
print(n - 1, m)
```
| 10,978 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
for _ in range(int(input())):
x,y = map(int,input().split(' '))
if x==1: print(0,y)
else: print(x-1,y)
```
| 10,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
mx=10**7
spf=[mx]*(mx+1)
def readTree(n,e): # to read tree
adj=[set() for i in range(n+1)]
for i in range(e):
u1,u2=IP()
adj[u1].add(u2)
return adj
def sieve():
li=[True]*(10**3+5)
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime,cur=[0]*200,0
for i in range(10**3+5):
if li[i]==True:
prime[cur]=i
cur+=1
return prime
def SPF():
mx=(10**6+1)
spf[1]=1
for i in range(2,mx):
if spf[i]==1e9:
spf[i]=i
for j in range(i*i,mx,i):
if i<spf[j]:
spf[j]=i
return
def prime(n,d):
prm=set()
while n!=1:
prm.add(spf[n])
n=n//spf[n]
for ele in prm:
d[ele]=d.get(ele,0)+1
return
#####################################################################################
mod=998244353
def solve():
x,y=IP()
print(x-1,y)
return
t=II()
for i in range(t):
solve()
#######
#
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```
| 10,980 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Tags: constructive algorithms, games, math
Correct Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
print(n-1,m)
```
| 10,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
# cook your dish here
from sys import stdin, stdout
import math
from itertools import permutations, combinations
from collections import defaultdict
from collections import Counter
from bisect import bisect_left
import sys
from queue import PriorityQueue
import operator as op
from functools import reduce
mod = 1000000007
def L():
return list(map(int, stdin.readline().split()))
def In():
return map(int, stdin.readline().split())
def I():
return int(stdin.readline())
def printIn(ob):
return stdout.write(str(ob) + '\n')
def powerLL(n, p):
result = 1
while (p):
if (p & 1):
result = result * n % mod
p = int(p / 2)
n = n * n % mod
return result
def ncr(n, r):
r = min(r, n - r)
numer = reduce(op.mul, range(n, n - r, -1), 1)
denom = reduce(op.mul, range(1, r + 1), 1)
return numer // denom
def SieveOfEratosthenes(n):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime_list = []
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
prime_list.append(p)
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime_list
def get_divisors(n):
for i in range(1, int(n / 2) + 1):
if n % i == 0:
yield i
yield n
# -------------------------------------
def myCode():
x,y = In()
print(x-1,y)
# print(d)
def main():
for t in range(I()):
myCode()
if __name__ == '__main__':
# print(lst)
main()
```
Yes
| 10,982 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
t=int(input())
for _ in range(t):
a,b=map(int,input().split())
print(a-1,b)
```
Yes
| 10,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
def program():
a,b=[int(x) for x in input().split()]
print(a-1,b)
if __name__ == "__main__":
test=1
test=int(input())
for i in range(test):
program()
```
Yes
| 10,984 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
from sys import stdin, stdout
ans = []
def main():
x,y = list(map(int,stdin.readline().split()))
ans.append(str(x-1)+' '+str(y))
if __name__ == '__main__':
for _ in range(int(stdin.readline())):
main()
# code same
stdout.write('\n'.join(ans))
# testing 1234567psdccv 3 dv
```
Yes
| 10,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
'''
Online Python Compiler.
Code, Compile, Run and Debug python program online.
Write your code in this editor and press "Run" button to execute it.
'''
n=int(input())
for i in range(n):
a,b=map(int,input().split())
if a<=b:
print(a-1,b)
else:
print((a-b),min(a,b))
```
No
| 10,986 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
t = int(input())
spis = []
for _ in range(t):
x1,y1 = 0,0
x, y = map(int, input().split())
if x < y:
if x > 1:
x1 = 1
y1 = y - (x - 1)
if x == 1:
x1 = 0
y1 = y
if x > y:
if y > 1:
x1 = x - y +1
y1 = 1
if y == 1:
x1 = x - 1
y1 = 1
if x == y:
if x == 1 and y == 1:
x1 = 0
y1 = 1
else:
x1 = 1
y1 = 1
spis.append([x1,y1])
for i in range(t):
for j in range(2):
print(spis[i][j],end=' ')
print()
```
No
| 10,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
from functools import reduce
import collections
import math
import sys
class Read:
@staticmethod
def string():
return input()
@staticmethod
def int():
return int(input())
@staticmethod
def list(delimiter=' '):
return input().split(delimiter)
@staticmethod
def list_int(delimiter=' '):
return list(map(int, input().split(delimiter)))
# infilename = 'input.txt'
# sys.stdin = open(infilename, 'r')
# outfilename = 'output.txt'
# sys.stdout = open(outfilename, 'w')
def main():
t = Read.int()
for _ in range(t):
a, b = Read.list_int()
if a == b:
print(a-1, a)
elif a - b == 1:
print(1, 1)
elif(a==1):
print(0,b)
else:
print(a, b)
if __name__ == '__main__':
main()
```
No
| 10,988 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play ping-pong with simplified rules.
During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.
The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.
Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.
Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.
Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.
Calculate the resulting number of Alice's and Bob's wins.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.
Output
For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.
Example
Input
3
1 1
2 1
1 7
Output
0 1
1 1
0 7
Note
In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.
In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.
In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
Submitted Solution:
```
for _ in range(int(input())):
x, y = map(int, input().split())
if y >= x:
print(0, y - x + 1)
else:
print(y, y)
```
No
| 10,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
t=int(input())
for _ in range(t):
a=str(input())
b=str(input())
x=len(a)
y=len(b)
if x>y:
mini=0
for i in range(y):
for j in range(i+1,y+1):
s=b[i:j]
l=j-i
p=0
for k in range(l,x+1):
if a[p:k]==s:
mini=max(mini,k-p)
p+=1
print(x+y-(2*mini))
else:
mini=0
for i in range(x):
for j in range(i+1,x+1):
s=a[i:j]
l=j-i
p=0
for k in range(l,y+1):
if b[p:k]==s:
mini=max(mini,k-p)
p+=1
print(x+y-(2*mini))
```
| 10,990 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
for i in range(int(input())):
a=list(input())
b=list(input())
a1=len(a)
b1=len(b)
if a1>b1:
a1,b1=b1,a1
a,b=b,a
s=[]
for i in range(a1):
for j in range(i+1,a1+1):
s.append(a[i:j])
s1=0
for i in range(b1):
for j in range(i+1,i+1+a1):
t=b[i:j]
if t in s:
s1=max(s1,len(t))
s1=a1+b1-s1-s1
print(s1)
```
| 10,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
def getCharNGram(n,s):
charGram = [''.join(s[i:i+n]) for i in range(len(s)-n+1)]
return charGram
t = int(input())
for _ in range(t):
A = input()
B = input()
if len(A) < len(B):
ans = len(B)+len(A)
for i in range(1,len(A)+1):
check = getCharNGram(i, A)
for c in check:
if c in B:
ans = min(ans, len(B)-i+len(A)-i)
else:
ans = len(A)+len(B)
for i in range(1,len(B)+1):
check = getCharNGram(i, B)
#print(check)
for c in check:
if c in A:
ans = min(ans, len(A)-len(c)+len(B)-i)
print(ans)
```
| 10,992 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
T = int(input())
for _ in range(T):
a = input()
b = input()
out = len(a) + len(b)
for i in range(len(a)):
for j in range(len(a), i, -1):
if a[i:j] in b:
out = min(out, len(a)+len(b)-2*(j-i))
print(out)
```
| 10,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
a = input().rstrip()
b = input().rstrip()
aa = set()
bb = set()
for i in range(len(a)):
c = []
for j in range(i, len(a)):
c.append(a[j])
aa.add(tuple(c))
for i in range(len(b)):
c = []
for j in range(i, len(b)):
c.append(b[j])
bb.add(tuple(c))
z = aa.intersection(bb)
max_len = 0
for i in z:
max_len = max(max_len, len(i))
ans = (len(a) + len(b)) - (max_len + max_len)
print(ans)
```
| 10,994 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
import sys,functools,collections,bisect,math,heapq
input = sys.stdin.readline
#print = sys.stdout.write
def fun(A ,B):
n = len(A)
m = len(B)
#Edge cases.
if m == 0 or n == 0:
return 0
if n == 1 and m == 1:
if A[0] == B[0]:
return 1
else:
return 0
#Initializing first row and column with 0 (for ease i intialized everthing 0 :p)
dp = [[0 for x in range(m + 1)] for y in range(n + 1)]
final = 0
#this code is a lot like longest common subsequence(only else condition is different).
for i in range(1, n + 1):
for j in range(1, m + 1):
if A[i - 1] == B[j - 1]:
dp[i][j] = 1 + dp[i - 1][j - 1]
else:
dp[i][j] = 0
final = max(final, dp[i][j])
return final
t = int(input())
for _ in range(t):
s = input().strip()
s1 = input().strip()
x = fun(s,s1)
print(len(s)+len(s1)-(2*x))
```
| 10,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
import sys
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def LS2(): return list(sys.stdin.readline().rstrip())
t = I()
for _ in range(t):
a = S()
b = S()
len_a = len(a)
len_b = len(b)
ans = len_a+len_b
for i in range(1,min(len_a,len_b)+1):
for j in range(len_a-i+1):
for k in range(len_b-i+1):
if a[j:j+i] == b[k:k+i]:
ans = min(ans,len_a+len_b-2*i)
print(ans)
```
| 10,996 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Tags: brute force, implementation, strings
Correct Solution:
```
n = int(input())
for i in range(n):
s1 = input()
s2 = input()
if (len(s1) > len(s2)): s1, s2 = s2, s1
max = 0
for j in range(len(s1)):
for k in range(j, len(s1)):
found = s2.find(s1[j:k+1])
if (found != -1):
if (k - j + 1 > max):
max = k - j + 1
else:
break
print(len(s1) + len(s2) - 2*max)
```
| 10,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
"""
"""
import sys
from sys import stdin
tt = int(stdin.readline())
ANS = []
for loop in range(tt):
a = list(stdin.readline()[:-1])
b = list(stdin.readline()[:-1])
ans = 0
for i in range(len(a)):
for j in range(len(b)):
now = 0
for k in range(min(len(a)-i,len(b)-j)):
if a[i+k] == b[j+k]:
now += 1
else:
break
ans = max(ans,now)
ANS.append(str(len(a)+len(b) - 2 * ans))
print ("\n".join(ANS))
```
Yes
| 10,998 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order:
* if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n;
* if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1};
* if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n;
* if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}.
Note that after each of the operations, the string a or b may become empty.
For example, if a="hello" and b="icpc", then you can apply the following sequence of operations:
* delete the first character of the string a ⇒ a="ello" and b="icpc";
* delete the first character of the string b ⇒ a="ello" and b="cpc";
* delete the first character of the string b ⇒ a="ello" and b="pc";
* delete the last character of the string a ⇒ a="ell" and b="pc";
* delete the last character of the string b ⇒ a="ell" and b="p".
For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.
Input
The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.
The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.
The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.
Output
For each test case, output the minimum number of operations that can make the strings a and b equal.
Example
Input
5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser
Output
0
2
13
3
20
Submitted Solution:
```
def getLCSSubStr(X, Y, m, n):
LCSuff = [[0 for i in range(n + 1)]
for j in range(m + 1)]
length = 0
row, col = 0, 0
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0:
LCSuff[i][j] = 0
elif X[i - 1] == Y[j - 1]:
LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1
if length < LCSuff[i][j]:
length = LCSuff[i][j]
row = i
col = j
else:
LCSuff[i][j] = 0
if length == 0:
return ""
resultStr = ['0'] * length
while LCSuff[row][col] != 0:
length -= 1
resultStr[length] = X[row - 1] # or Y[col-1]
row -= 1
col -= 1
return (''.join(resultStr))
def find_ans(X, Y):
m = len(X)
n = len(Y)
common_string = getLCSSubStr(X, Y, m, n)
a = m - len(common_string)
b = n - len(common_string)
return a + b
test_cases = int(input())
ans = []
for _ in range(test_cases):
X = input()
Y = input()
ans.append(find_ans(X, Y))
for i in ans:
print(i)
```
Yes
| 10,999 |
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